Question

Given a uniform disc of mass $M$ and radius $R$. A small disc of radius $R/2$ is cut from this disc in such a way that the distance between the centres of the two disc is $R/2$. Find the moment of inertia of the remaining disc about a diameter of the original disc perpendicular to the line connecting the centres of the two discs

• A. $3M{R}^2/32$
• B. $5M{R}^2/16$
• C. $11M{R}^2/64$
• D. none of these

Answer 1

The correct option is C $11M{R}^2/64$

Mass of the cut-out ring$M^{\prime }=\frac{M}{4}$ (As radius is half so the area is one-fourth)

Moment of inertia of original ring about an axis passing through its centre and lying in the ring plane = $\frac{M{R}^2}{4}$

Moment of inertia of cut out a ring about an axis passing through its centre and in the plane of the ring $I_{cut}=M^{\prime }\frac{(\frac{R}{2}{)}^2}{4}+M^{\prime }(\frac{R}{2}{)}^2$

$I_{balance}=\frac{M{R}^2}{4}-I_{cut}=\frac{M{R}^2}{4}-M^{\prime }\frac{(\frac{R}{2}{)}^2}{4}-M^{\prime }(\frac{R}{2}{)}^2=\frac{M{R}^2}{4}-I_{cut}$

$=\frac{M{R}^2}{4}-\frac{M}{4}\frac{(\frac{R}{2}{)}^2}{4}-\frac{M}{4}(\frac{R}{2}{)}^2=\frac{11M{R}^2}{64}$